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Chapter 20: Problem 65

A volume of \(6.00 \mathrm{~L}\) of a monatomic ideal gas, originally at \(400 . \mathrm{K}\) and a pressure of \(3.00 \mathrm{~atm}\) (called state 1 ), undergoes the following processes, all done reversibly: \(1 \rightarrow 2\) isothermal expansion to \(V_{2}=4 V_{1}\) \(2 \rightarrow 3\) isobaric compression \(3 \rightarrow 1\) adiabatic compression to its original state Find the entropy change for each process.

Short Answer

Expert verified
Question: Calculate the entropy change for each process that a monatomic ideal gas undergoes during the following three process cycle: 1) Isothermal expansion from V1 to V2 = 4 * V1, 2) Isobaric compression, and 3) Adiabatic compression back to the original state. The initial conditions of the gas are P1 = 3.00 atm, V1 = 6.00 L, and T1 = 400 K. Answer: The entropy changes for each process are given by: 1) Isothermal expansion: ΔS₁→₂ = nR ln(4) 2) Isobaric compression: ΔS₂→₃ = n(5/2)R ln(T₂/T₁) 3) Adiabatic compression: ΔS₃→₁ = 0

Step by step solution

01

Isothermal expansion

To find the entropy change in an isothermal process, we can use the formula: \(\Delta S = nR\ln \left(\frac{V_2}{V_1}\right)\) We first need to find the number of moles, n, using the initial conditions of the gas: \(P_1 = 3.00\ \mathrm{atm},\ V_1 = 6.00\ \mathrm{L},\ T_1 = 400\ \mathrm{K}\) Using the ideal gas law \(PV=nRT\), we have: \(n = \frac{P_1V_1}{RT_1} = \frac{3.00\ \mathrm{atm} \cdot 6.00\ \mathrm{L}}{0.0821\ \mathrm{L\ atm/mol\ K} \cdot 400\ \mathrm{K}}\) Now, we can find the entropy change in the isothermal expansion: \(\Delta S_{1\rightarrow2} = nR\ln \left(\frac{V_2}{V_1}\right) = nR\ln(4)\)
02

Isobaric compression

For an isobaric process, the formula for the entropy change is: \(\Delta S = nC_P\ln \left(\frac{T_2}{T_1}\right)\) Since the process is isobaric, the pressure at state 2 is the same as state 1. We can use the ideal gas law to determine the temperature at state 2: \(P_1V_1 = nRT_1 = nRT_2\) Rearranging for T2 gives: \(T_2 = \frac{P_1V_1}{nR} = \frac{3.00\ \mathrm{atm} \cdot 6.00\ \mathrm{L}}{0.0821\ \mathrm{L\ atm/mol\ K}}\) Because the gas is monatomic, we have: \(C_P = \frac{5}{2}R\) Now, we can find the entropy change in the isobaric compression: \(\Delta S_{2\rightarrow3} = nC_P\ln \left(\frac{T_2}{T_1}\right) = n\left(\frac{5}{2}R\right)\ln\left(\frac{T_2}{T_1}\right)\)
03

Adiabatic compression

In an adiabatic process, the entropy change is \(0\) since there is no heat transfer: \(\Delta S_{3\rightarrow1} = 0\) Finally, we calculate the entropy change for each process using the equations derived in the previous steps: \(\Delta S_{1\rightarrow2} = nR\ln(4)\) \(\Delta S_{2\rightarrow3} = n\left(\frac{5}{2}R\right)\ln\left(\frac{T_2}{T_1}\right)\) \(\Delta S_{3\rightarrow1} = 0\) Using these entropy changes, we are able to describe the entropy changes in the three processes which the monatomic ideal gas undergoes.

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Most popular questions from this chapter

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The entropy of a macroscopic state is given by \(S=k_{B} \ln w\) where \(k_{\mathrm{B}}\) is the Boltzmann constant and \(w\) is the number of possible microscopic states. Calculate the change in entropy when \(n\) moles of an ideal gas undergo free expansion to fill the entire volume of a box after a barrier between the two halves of the box is removed.

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